For simplicity, we write \(\sigma \)  instead
of \(\sigma (\cdot , \, a, \, v(a))\) .
Recalling the definition of \(\sigma \) , and writing the Green function
in the form (REF ), we obtain
\( \sigma (x) = -\frac{1}{2\pi } \left(\nabla _a \log \operatorname{dist}_g(x, \, a),\, v(a)\right)_g+ \left(\nabla _a H(x, \, a), \, v(a)\right)_g\) 
The term \((\nabla _a H(x, \, a), \, v(a))_g\)  is \(\mathrm {O}(1)\) 
because \(H\)  is of class \(C^1\)  in both variables.
We compute \(\nabla _a\log \operatorname{dist}_g(x, \, a)\) .
By Gauss' lemma, \(\nabla _a\operatorname{dist}_g(x, \, a) = \tilde{\nu }(a)\in T_a M\) ,
where \(\tilde{\nu }\)  is the unit vector field
that is orthogonal to the geodesic circles centered at \(x\) 
and points outward. By the chain rule, we obtain
\( \begin{split}\sigma (x) = -\frac{1}{2\pi \operatorname{dist}_g(x, \, a)}\left(\tilde{\nu }(a), \, v(a)\right)_g+ \left(\nabla _a H(x, \, a), \, v(a)\right)_g\end{split}\) 
If \(x\)  and \(a\)  are close enough, there exists
a unique minimizing geodesic \(\gamma \colon [0, \, 1]\rightarrow M\) 
such that \(\gamma (0)=a\) , \(\gamma (1) =x\) .
Let \(\tilde{\gamma }\colon [0, \, 1]\rightarrow M\) 
be the geodesic defined by \(\tilde{\gamma }(t) := \gamma (1-t)\) .
Let \(P_{x,a}\colon T_x M \rightarrow T_a M\)  denote parallel
transport along \(\tilde{\gamma }\)  (see e.g. [1]}).
Then,
\(\tilde{\nu }(a) = \tilde{\gamma }^\prime (1) = P_{x,a}\tilde{\gamma }^\prime (0)= - P_{x,a}\gamma ^\prime (1) = -P_{x,a}\nu (x)\) 
and hence, (REF ) can be written as
\( \begin{split}\sigma (x)&= \frac{1}{2\pi \operatorname{dist}_g(x, \, a)}\left(P_{x,a}\nu (x), \, v(a)\right)_g+ \left(\nabla _a H(x, \, a), \, v(a)\right)_g\end{split}\) 
The parallel transport map \(P_{x,a}\)  is an isometry,
and its inverse \(P_{a,x} := P_{x,a}^{-1}\colon T_a M\rightarrow T_x M\) 
is the parallel tranport along \(\gamma \) . Then, (REF )
is equivalent to
\( \begin{split}\sigma (x)&= \frac{1}{2\pi \operatorname{dist}_g(x, \, a)}\left(\nu (x), \, P_{a,x}v(a)\right)_g+ \left(\nabla _a H(x, \, a), \, v(a)\right)_g \ \end{split}\) 
Since both \(v\)  and \(x\mapsto P_{a,x}v(a)\)  are smooth,
with \(P_{a,a}v(a) = v(a)\) , we have
\( \left| P_{a,x}v(a) - v(x) \right|_g = \mathrm {O}(\operatorname{dist}_g(x, \, a))\) 
and hence, (REF ) follows.
We pass to the proof of ().
By differentiating (REF ), we obtain
\( \begin{split}\mathrm {d}\sigma (x) &= \frac{1}{2\pi }\mathrm {d}\left(\frac{\nu (x)}{\operatorname{dist}_g(x, \, a)}, \, P_{a,x}v(a)\right)_g+ \mathrm {d}\left(\nabla _a H(x, \, a), \, v(a)\right)_g\end{split}\) 
Here \(\mathrm {d}\)  denotes the exterior differential with respect to \(x\) .
From now on, we treat \(a\)  as a fixed parameter.
By Lemma REF , \(\mathrm {d}(\nabla _a H(x, \, a), \, v(a))_g =\mathrm {O}(\log \operatorname{dist}_g(x, \, a))\) . Moreover, the vector
field \(x\mapsto P_{a,x} v(a)\)  is smooth (in particular,
its derivatives are bounded). Therefore,
\( \begin{split}\mathrm {d}\sigma (x) &= -\frac{(\nu (x), \, P_{a,x}v(a))_g}{2\pi \operatorname{dist}_g(x, \, a)^2}\, \mathrm {d}\left(\operatorname{dist}_g(x, \, a)\right)+ \frac{(D \nu (x), \, P_{a,x} v(a))}{2\pi \operatorname{dist}_g(x, \, a)}+ \mathrm {O}(\operatorname{dist}_g(x, \, a)^{-1}) \\&= -\frac{(\nu (x), \, P_{a,x}v(a))_g}{2\pi \operatorname{dist}_g(x, \, a)^2} \, \nu ^\flat (x)+ \frac{(i\nu (x), \, P_{a,x} v(a))_g}{2\pi \operatorname{dist}_g(x, \, a)^2} (i\nu (x))^\flat + \mathrm {O}(\operatorname{dist}_g(x, \, a)^{-1})\end{split}\) 
The second equality follows by Gauss' lemma
and Lemma REF .
Finally, by applying (REF ) and Lemma REF ,
we deduce
\( \begin{split}\mathrm {d}\sigma (x) &= -\frac{(\nu (x), \, v(x))_g}{2\pi \operatorname{dist}_g(x, \, a)^2} \, \nu ^\flat (x)+ \frac{(i\nu (x), \, v(x))_g}{2\pi \operatorname{dist}_g(x, \, a)^2} \star \nu ^\flat (x)+ \mathrm {O}(\operatorname{dist}_g(x, \, a)^{-1})\end{split}\) 
and the lemma follows.
